A 4.50 g mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end of the spring is fixed to a wall. It takes 3.6J of work to compress the spring by 13.0 cm. If the spring is compressed, and the mass is released from rest. find a) the period of vibration, and b) the velocity of the mass when its is 5.00 cm from the equilibrium.
m=0.0045kg
E=3.6 J for .13m compression
T=?
Vm @ .05m from equilibrium= ?
P.E.= .5KA^2
therefore find K=F/x
(.0045x9.8)/(.13)
K =.34N/m
T=2pixsprt(m/K)
T= 2xpi x sprt (.0045/.34)
T= .72 secs
B) E=PE+KE
3.6= .5 (3.4)(0)^2 + .2 (.0045)(v^2)
v^2=1600
v=40
To find the period of vibration and the velocity of the mass, we can use the principles of energy conservation and simple harmonic motion.
a) The period of vibration can be found using the formula:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
To find the spring constant, we can use the formula:
PE = (1/2)kx^2
where PE is the potential energy stored in the spring, x is the compression distance, and k is the spring constant.
Given that it takes 3.6J of work to compress the spring by 13.0 cm, we can calculate the potential energy:
PE = 3.6 J
Thus, using the formula for potential energy, we have:
(1/2)k(0.13 m)^2 = 3.6 J
k(0.0169 m^2) = 3.6 J
k = 3.6 J / 0.0169 m^2
k ≈ 213.02 N/m
Now, we can find the period:
T = 2π√(m/k)
T = 2π√(4.50 g / (213.02 N/m))
T = 2π√(0.00450 kg / 213.02 N/m)
T ≈ 0.436 s (rounded to three decimal places)
b) To find the velocity when the mass is 5.00 cm from the equilibrium, we need to find the maximum displacement from equilibrium, amplitude (A).
The amplitude is equal to the initial compression distance:
A = 0.13 m
Since we know the amplitude, we can use the formula for simple harmonic motion:
v = ωA
where v is the velocity, ω is the angular frequency, and A is the amplitude.
The angular frequency can be calculated using:
ω = 2π / T
Given T = 0.436 s, we find:
ω = 2π / 0.436 s
ω ≈ 14.413 rad/s (rounded to three decimal places)
Now we can find the velocity when the mass is 5.00 cm from the equilibrium:
v = ωA
v = 14.413 rad/s * 0.05 m
v ≈ 0.721 m/s (rounded to three decimal places)
Therefore, the period of vibration is approximately 0.436 s, and the velocity of the mass when it is 5.00 cm from the equilibrium is approximately 0.721 m/s.